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Math Help - Help on limit at infinite problem?

  1. #1
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    Help on limit at infinite problem?

    Lim [sqrt (x^2 + 3x)] / (4-4x)
    x> -inf

    -inf means -infinity and its the square root of only (x^2 + 3x)

    How do I even do a problem like this? I found -1/4 which is wrong
    according to my webwork site.
    Then I found -inf which is also wrong since it sais the point im finding will be the horizontal asymptote.
    Please help!
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  2. #2
    MHF Contributor Calculus26's Avatar
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    Just like with rational fns consider the ratio of highest powered terms


    Lim [sqrt (x^2 + 3x)] / (4-4x)
    x> -inf

    Recall sqrt(x^2) = |x| = -x for x <0 since x-> negative inf we have

    = lim |x|/(-4x) = lim-x/-4x = 1/4
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